A balloon is pumped at the rate of a $\dfrac{\text{c}{{\text{m}}^{\text{3}}}}{\text{minute}}$ . The rate of increase of its surface area when the radius is b cm, is
(a) $\dfrac{2{{a}^{2}}}{{{b}^{4}}}\dfrac{\text{c}{{\text{m}}^{2}}}{\text{min}}$
(b) $\dfrac{a}{2b}\dfrac{\text{c}{{\text{m}}^{2}}}{\text{min}}$
(c) $\dfrac{2a}{b}\dfrac{\text{c}{{\text{m}}^{2}}}{\text{min}}$
(d) None of these

Hint: First, we have to identify which data is given to us. So, we are given with $\dfrac{dV}{dt}=a$ . Then we will use formula of volume of sphere and will differentiate with respect to time as we can see the unit is in $\dfrac{\text{c}{{\text{m}}^{\text{3}}}}{\text{minute}}$ . Then from this we will get the value $\dfrac{dr}{dt}$ and then substitute this value after differentiating the surface area given as $S=4\pi {{r}^{2}}$ . So, We will get the final answer in the form of $\dfrac{dS}{dt}$ .
Formula for differentiating will be $\dfrac{d}{dt}\left( {{x}^{2}} \right)=2x\dfrac{dx}{dt}$ .

Complete step-by-step solution -
Here, we have to find the rate of increase of its surface area when radius of balloon is b cm and it is also given that balloon is pumped at rate of a $\dfrac{\text{c}{{\text{m}}^{\text{3}}}}{\text{minute}}$ . We will consider the balloon as a shape of sphere and then will solve the problem.
So, here we will first use the formula of Volume of sphere i.e. given as $\dfrac{4}{3}\pi {{r}^{3}}$ .
$\therefore Volume\left( V \right)=\dfrac{4}{3}\pi {{r}^{3}}$
Now, we will differentiate Volume with respect to time by using formula of differentiation i.e. for example $\dfrac{d}{dt}\left( {{x}^{2}} \right)=2x\dfrac{dx}{dt}$ . so, we will get
$\therefore \dfrac{dV}{dt}=\dfrac{4}{3}\pi \dfrac{d}{dt}\left( {{r}^{3}} \right)$
On solving, we get
$\therefore \dfrac{dV}{dt}=\dfrac{4}{3}\pi \cdot 3{{r}^{2}}\dfrac{dr}{dt}$
Cancelling 3 on RHS side, we get
$\therefore \dfrac{dV}{dt}=4\pi {{r}^{2}}\dfrac{dr}{dt}$
Here, we are given that $\dfrac{dV}{dt}=a$ and radius has become b cm. So, replacing r as b and after substituting the values, we get
$\therefore a=4\pi {{b}^{2}}\dfrac{dr}{dt}$
On taking constant term on LHS, we get
$\therefore \dfrac{a}{4\pi {{b}^{2}}}=\dfrac{dr}{dt}$ ………………………………………(1)
Now, we have to find rate on increase surface area by using the formula $S=4\pi {{r}^{2}}$
So, again differentiating the above formula, we get
$\dfrac{dS}{dt}=4\pi \dfrac{d}{dt}\left( {{r}^{2}} \right)$
Using the differentiation formula $\dfrac{d}{dt}\left( {{x}^{2}} \right)=2x\dfrac{dx}{dt}$ , we get
$\Rightarrow \dfrac{dS}{dt}=4\pi 2r\dfrac{dr}{dt}$
$\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$
Now, substituting the value of equation (1) and putting radius r as b, we get
$\Rightarrow \dfrac{dS}{dt}=8\pi b\dfrac{a}{4\pi {{b}^{2}}}$
On simplification, we get
$\Rightarrow \dfrac{dS}{dt}=\dfrac{2a}{b}$
Thus, the rate of increase of its surface area when the radius is b cm, is $\dfrac{2a}{b}\dfrac{\text{c}{{\text{m}}^{2}}}{\text{min}}$
Hence, option (c) is correct.

Note: Be sure while differentiating with respect to time variables. Students make mistake in differentiating radius r variable and forget to put $\dfrac{dr}{dt}$ i.e. $\dfrac{dV}{dt}=\dfrac{4}{3}\pi \cdot 3{{r}^{2}}=4\pi {{r}^{2}}$ . This value will be the direct formula of surface area of sphere. On doing further differentiation, we will get the answer as $8\pi b$ on putting r as b and answer will be completely wrong. So, don’t forget to put $\dfrac{dr}{dt}$ which is very important in this problem.